Probability of Balanced Trial with Maximum Excess in 3-Treatment Experiment

[_joey]

Given 10 strata. 3 blocks in each strata and 3 treatments.

Question: what is the probability that the trial will end exactly balanced for 2 of the 3 treatments and with the maximum excess for the other treatment?

My solution about which I am unsure.

The trial is exactly balanced fro 2 of the 3 treatments and with maximum excess for the other treatment when each stratum is 1/3 through the block.

The probability of stopping 1/3 through the block for 10 strata is (1/3)^10.
The probability that the first treatment in the block is the same treatment across all strata is also (1/3)^10.
There are 3 treatments and each trial could end in favour any of the 3 treatments.

So, the probability is 3*(1/3)^10*(1/3)^10.

What do you think?

Thanks!

 

[haruspex]

There are many on this forum who could help you with the probability aspect but are ignorant of experimental design terminology. Can you re-post this in layman's language?

 

[_joey]

There're 10 queue lines. Person entering each queue line is assigned a certain identifier. There are only 3 identifiers let say A, B, C. People entering 10 lines are divided in groups of 3 such that each group always have balanced number of people with different identifiers. For example, ABC, ACB, BCA etc but not AAB or CCC. Assume people are entering the lines randomly and uniformly and lines are independent of each other.

Question: what is the probability that the lines will end balanced for any 2 of the 3 identifiers and with the maximum excess for the other identifier?

 

[_joey]

There are many on this forum who could help you with the probability aspect but are ignorant of experimental design terminology. Can you re-post this in layman's language?

Although, I identified the problem as simple probability problem, I don't think it fits 'pre-calculus level maths'.

 

[haruspex]

There're 10 queue lines. Person entering each queue line is assigned a certain identifier. There are only 3 identifiers let say A, B, C. People entering 10 lines are divided in groups of 3 such that each group always have balanced number of people with different identifiers. For example, ABC, ACB, BCA etc but not AAB or CCC. Assume people are entering the lines randomly and uniformly and lines are independent of each other.

I assume we're not to worry about people leaving the queues - they just get longer.
Not quite clear about the groups of three. Can a queue consist of A ABC CBA BAC (front)?
What determines the assignments? Is it random and equally likely apart from the no repeats rule?

Question: what is the probability that the lines will end balanced for any 2 of the 3 identifiers and with the maximum excess for the other identifier?

Is this still treating the lines separately? If so, how is the number of lines interesting?
Is it equally likely that the final queue length is 0, 1 or 2 mod 3?

 

[_joey]

I assume we're not to worry about people leaving the queues - they just get longer.
Not quite clear about the groups of three. Can a queue consist of A ABC CBA BAC (front)?
What determines the assignments? Is it random and equally likely apart from the no repeats rule?

People can enter as ABC, CBA, BAC.
The assignment maybe determined by sex, age or martial status.

Is this still treating the lines separately? If so, how is the number of lines interesting?
Is it equally likely that the final queue length is 0, 1 or 2 mod 3?

It's equally likely. The number of lines are interesting as we searching for the probability that all lines will be balanced given the constraints.

 

[mfb]

People can enter as ABC, CBA, BAC.
The assignment maybe determined by sex, age or martial status.

So they just enter in groups of 3?

It's equally likely. The number of lines are interesting as we searching for the probability that all lines will be balanced given the constraints.

If all lines are independent of each other, we can focus on one line. If they are not independent, how does the dependency look like?

 

[_joey]

^The lines are independent. The solution is given above for commenting.

 

[mfb]

If we think of the same assumptions now, your formula in post 1 is right.

So "maximum excess" is 1? Why don't you call it 1 then?

 

[haruspex]

I don't think you're being entirely consistent. You started off with

exactly balanced fro 2 of the 3 treatments and with maximum excess for the other treatment when each stratum is 1/3 through the block.

If a stratum is 1/3 of the way through a block then necessarily there is an excess of 1 for one treatment, so the answer to that is 1 (certainty).
Next, you wrote:

the probability that the lines will end balanced for any 2 of the 3 identifiers and with the maximum excess for the other identifier

That is the same as asking the probability that the line ends 1/3 of the way through a block, giving a probability of 1/3. The probability that all lines will be in this state will be (1/3)¹⁰, but then you wrote:

we searching for the probability that all lines will be balanced

So that means we don't care whether a line has an excess of one for one treatment or a deficit of one for one treatment. The probability that a given line is balanced is 1/3, etc.
The formula you originally posted looks about right if you are interested in the case where the same treatment is one in excess in all lines.

 

[_joey]

I don't think you're being entirely consistent. You started off with

If a stratum is 1/3 of the way through a block then necessarily there is an excess of 1 for one treatment, so the answer to that is 1 (certainty).
Next, you wrote:

That is the same as asking the probability that the line ends 1/3 of the way through a block, giving a probability of 1/3. The probability that all lines will be in this state will be (1/3)¹⁰, but then you wrote:

So that means we don't care whether a line has an excess of one for one treatment or a deficit of one for one treatment. The probability that a given line is balanced is 1/3, etc.
The formula you originally posted looks about right if you are interested in the case where the same treatment is one in excess in all lines.

The example is from the textbook. I feel like explaining to the students what the problem asks about. Do you know much about probabilities?

 

[haruspex]

Let's just think about one line first, to clarify things.
Do you want the probability:
a) that the line ends balanced?
b) that the line ends imbalanced with one extra occurrence of one treatment (we don't care which)?
c) that the line ends imbalanced with one extra occurrence of a particular treatment?

With multiple lines, do you want the probability that:
x) all lines end balanced?
y) all lines end imbalanced as in (b)?
z) all lines end imbalanced as in (c), with the same extra treatment in each case?

 

[_joey]

^ No disrespect, I don't think I need your help. :)